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TOPIC 1 : RATES OF REACTIONS
RATE OF REACTION
The rate of chemical reaction measures how quickly a reactant is used up, or a product is
formed. It is defined as the change in the amount of a reactant or product per unit time.
THE COLLISION THEORY
For a chemical reaction to occur;
the reactant particles must collide with each other
the particles must have enough energy for the collision to be successful
A successful collision is the one that produces a reaction. The rate of reaction depends on
how many successful collisions there are in a given time. The more successful collisions
there are, the faster the reaction. The minimum amount of energy needed for a collision to be
successful is called activation energy.
FACTORS AFFECTING THE RATE OF REACTION
Some of the factors that affect the rate of a reaction are:
a. Concentration of the reactants
The rate of a reaction increases when the concentration of a reactant is increased. When the
concentration of the solution increases, the amount of the particles per unit volume also
increases. Frequency of collisions between reacting particles increases.
b. Temperature
The rate of reaction increases with an increase in temperature. At higher temperatures, the
reactant particles have more kinetic energy, they move faster, and collide more often with
greater energy causing the rate of reaction to increase.
c. Surface area of solid reactants
When the size of particles of a solid decrease, surface area of the solid increases, allowing
them to collide with other reacting particles more readily hence the rate of reaction of
increases. This is why powdered solids react faster than lumps.
d. Use of a catalyst
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A catalyst is a substance that speeds up a chemical reaction, but remains chemically
unchanged itself. Catalysts work by lowering the activation energy for the reaction. Lowering
the activation energy increases the frequency of successful collisions because a greater
proportion of collisions now exceed this lower energy.
METHODS OF MEASURING RATES OF REACTIONS
There are different methods to measure the rate of reaction. The method chosen usually
depends on the reactants and products involved and how easy it is to measure the changes in
them.
a. Measuring the volume of gas produced.
b. Measuring the change in mass of reactants or products.
c. Measuring the time taken to for a given mass of reactant to disappear completely.
d. Measuring the time taken for a certain amount of product to form.
MEASURING THE VOLUME OF A GAS PRODUCED.
This method is used when a gas leaves the container. The volume is measured using a gas
syringe.
The volume of the gas given off is measured at regular time intervals, for example every half
a minute. Using the syringe, the volume of the gas give off, is noted by checking how the
plunger has been pushed. For example, if the plunger is pushed to the 20 cm
3
mark, it means
20 cm
3
of gas has been collected.
Instantaneous rate of reaction is calculated by the following formula:
Rate of reaction = volume of gas at
2
t
volume of gas at
1
t
2
t
1
t
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Average rate of reaction is calculated by the following formula:
Average rate of reaction = total volume of gas produced
reaction time
Worked example
In an experiment, some magnesium and excess of dilute hydrochloric acid were reacted
together. The volume of hydrogen produced was recorded every half a minute.
Time (minutes) Volume of hydrogen (cm
3
)
0 0
0.5 8
1.0 14
1.5 20
2.0 25
2.5 29
3.0 33
3.5 36
4.0 38
4.5 39
5.0 40
5.5 40
6.0 40
a. Plot a graph of volume of hydrogen against time.
b. What is the rate of reaction (in cm
3
of hydrogen per minute) during, the 1
st
minute, the 2
nd
minute and the 3
rd
minute?
c. How much hydrogen was produced in total?
d. How long does the reaction last?
e. Calculate the average rate of reaction
Solutions
a. The graph of volume of hydrogen against time is shown below
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b. Rate of reaction in the 1
st
minute = 14 cm
3
of hydrogen per minute
Rate of reaction in the 2
nd
minute =
25 14
1
= 11cm
3
of hydrogen per minute
Rate of reaction in the 3
rd
minute =
33 25
1
= 8cm
3
of hydrogen per minute
c. In total, 40 cm
3
of hydrogen was produced.
d. After 5 minutes no more hydrogen is produced. The reaction lasts 5 minutes
e. Average rate of reaction = total volume of hydrogen
total time for the reaction
=
3
40cm
5minutes
= 8cm
3
of hydrogen per minute
REVERSIBLE AND IRREVERSIBLE REACTIONS
a. Reversible reaction
A reversible reaction is a chemical reaction in which the products can be converted back to
reactants under suitable conditions.
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Example
When blue copper hydrated copper (II) sulphate is heated, it loses it water of crystallization.
A white anhydrous copper sulphate is formed. The equation for the dehydration of blue
copper sulphate is:
CuSO
4
.5H
2
O (s) CuSO
4
(s) + 5H
2
O (l)
(blue crystals) (white powder)
When water is added to the white anhydrous copper (II) sulphate powder, the colour turns
back to blue. This is because the water of crystallization is regained. The equation for
hydration of white anhydrous copper sulphate is:
CuSO
4
(s) + 5H
2
O (l) CuSO
4
.5H
2
O (s)
(white powder) (blue crystals)
The overall reaction is given as
CuSO
4
.5H
2
O (s) CuSO
4
(s) + 5H
2
O (l)
b. Irreversible reaction
An irreversible reaction is a reaction in which products cannot be converted back into
reactants.
Example
When calcium carbonate is decomposed, calcium oxide and carbon dioxide are produced.
The reaction cannot be reversed. The carbon dioxide escapes into the air.
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TOPIC 2 : ACIDS AND BASES
THE BRØNSTED-LOWRY THEORY OF ACIDS AND BASES
The BrØnsted-Lowry theory of acids and bases was proposed by a Danish chemist, Johnnes
BrØnsted and an English chemist, Thomas Lowry. Their definition centres on the proton, H
+
.
The proton is what remains after the normal hydrogen atom loses an electron. According to
the BrØnsted-Lowry theory:
An acid is a proton donor.
HCl H
+
+
-
Cl
Acid Proton
A base is a proton acceptor.
H
2
O + H
+
H
3
O
+
Base Proton
The overall acid-base reaction is: HCl + H
2
O H
3
O
+
+
-
Cl
Acid Base
FORMATION OF THE HYDRONIUM ION, H
3
O
+
The hydronium ion is formed when an acid dissolves in water, and the water gains the
hydrogen ions which are donated by the acid.
For example, when hydrochloric acid dissolves in water, it dissociates into hydrogen ions, H
+
and chloride ions,
-
Cl
as follows.
HCl H
+
+
-
Cl
The protons are then gained by the water to form the hydronium ion, H
3
O
+
as follows:
HCl + H
2
O H
3
O
+
+
-
Cl
CONJUGATE ACIDS AND CONJUGATE BASES
The terms conjugate acid and conjugate base are used to decribe products of acid-base
reactions.
A conjugate acid is the ion that is formed when a base gains a proton. Examples are:
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a. NH
3
+ H
+
+
4
NH
Base Proton Conjugate acid
b.
-
OH
+ H
+
H
2
O
Base Proton Conjugate acid
The ion formed is called the conjagate acid because, if we imagine the reverse of the reaction,
it can give up a proton.
A conjugate base is the ion that is formed when an acid loses a proton. Examples are:
a. CH
3
COOH
-
3
CH COO
+ H
+
Acid Conjugate base Proton
b.
+
4
NH
NH
3
+ H
+
Acid Conjugate base Proton
The ion formed is called the conjugate base because if we imagine, the reverse of the
reaction, it can accept a proton.
Examples
Identifying conjugate acid and conjugate base in acid-base reactions
a. CH
3
COOH + HCl CH
3
COOH
2
+
-
Cl
Base Acid Conjugate acid Conjugate base
b. HNO
3
+ H
2
SO
4
+
2 3
H NO
+
-
4
HSO
Base Acid Conjugate acid Conjugate base
c. HCl + HF
-
Cl
+ H
2
F
+
Acid Base Conjugate base Conjugate acid
CONJUGATE ACID-BASE PAIRS
A conjugate acid-base pair consists of molecules or ions which are connected by the loss of
one proton, H
+
by an acid and by the gain of one proton, H
+
by a base.
Every acid-base reactions contains two conjugate acid-base pairs, because the H
+
is
transferred in both the forward and reverse reactions.
Examples
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Write down the conjugate acid-base pairs in the following equations.
a. NH
3
(g) + H
2
O (l)
+
4
NH
(aq) +
-
OH
(aq)
Solution
The conjugate acid-base pairs are
+
4
NH
(aq)/NH
3
(g) and H
2
O (l)/
-
OH
(aq)
b. HNO
3
+ H
2
SO
4
+
2 3
H NO
+
-
4
HSO
Solution
The conjugate acid-base pairs are: H
2
SO
4
/
-
4
HSO
and
+
2 3
H NO
/ HNO
3
c. CH
3
COOH (l) + H
2
O (l)
-
3
CH COO
(aq) + H
3
O
+
(aq)
Solution
The conjugate acid-base pairs are: CH
3
COOH (l)/
-
3
CH COO
(aq) and H
3
O
+
(aq)/H
2
O (l)
STRENGTH OF ACIDS AND BASES
The strength of a particular acid or base is based on the degree to which it dissociate into ions
in aqueous solutions.
STRONG AND WEAK ACIDS
A strong acid is the one which ionizes completely in water to give all its hydrogen ions.
Examples of strong acids are hydrochloric acid ( HCl), sulphuric acid (H
2
SO
4
) and nitiric
acid (HNO
3
), etc.
The ionization equations for the acids are shown below:
a. HCl (l) + H
2
O (l) H
+
(aq) +
-
Cl
(aq)
b. H
2
SO
4
(l) + H
2
O (l) 2H
+
(aq) +
2-
4
SO
(aq)
c. HNO
3
(l) + H
2
O (l) H
+
(aq) +
-
3
NO
(aq)
A weak acid is the one which ionizes partially in water to produce few hydrogen ions.
Examples of weak acids are ethanoic acid (CH
3
COOH), carbonic acid (H
2
CO
3
) etc.
The ionisation equations for the acids are shown below:
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a. CH
3
COOH (l) + H
2
O (l)
-
3
CH COO
(aq) + H
3
O
+
(l)
b. H
2
CO
3
(l) + H
2
O (l)
-
3
HCO
(aq) + H
3
O
+
(l)
The double arrow
means that the reaction is reversible.
STRONG AND WEAK BASES
A strong base is the one which ionizes completely in water to produce more hydroxide ions.
Examples of stong bases are sodium hydroxide and potassium hydroxide.
The ionization equations for the bases are:
a. NaOH (s) + H
2
O (l) Na
+
(aq) +
-
OH
(aq)
b. KOH (s) + H
2
O (l) K
+
(aq) +
-
OH
(aq)
A weak base is the one which ionizes partially in water to produce few hydroxide ions. A
typical example of a weak base is ammonia.
NH
3
(g) + H
2
O (l)
+
4
NH
(aq) +
-
OH
(aq)
DETERMINING THE STRENGTH OF ACIDS AND BASES USING
CONDUCITIVITY APPARATUS
Acids and bases are able to conduct electricity because they ionize when dissolved in water.
However, the electrical conductivirty depends on the acid or base being used as electrolyte.
Stong bases and acids have higher electrical conductivity as compared to weak acids and
bases. This is because stong acids and bases ionize completely releasing all the ions which
conduct electricity when dissolvee in water.
On the other hand, weak acids and bases are weak electrolytes because they do not ionize
completely when dissolve din water and therefore release few ions to conduct electricity.
COMPARING THE ELECTRICAL CONDUCTIVITY OF STRONG AND WEAK
ACIDS AND BASES.
Materials: a beaker, 2 carbon rods, 2 cells, connecting wires, distilled water, ammeter,
solutions of ethanoic acid, hydrochloric acid, sodium hydroxide and ammonia (of equal
concentration)
Procedure:
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a. Put 50 ml of ethanoic acid solution into the beaker.
b. Arrange the apparatus as shown below.
c. Close the switch and record the ammeter reading in the table of results.
d. Rinse the beaker and the carbon electrodes with the distilled water.
e. Repeat steps a to d using 50 ml of the solutions of hydrochloric acid, sodium hydroxide
and ammonia respectively.
Table of results
Solution Ammeter reading
Ethanoic acid
Hydrochloric acid
Sodium hydroxide
Ammonia
Conclusion
The acid or base whose current reading is higher is a strong acid or base, and the acid or base
whose current reading is lower is a weak acid or base.
CONTROL VARIABLES DURING CONDUCTIVITY MEASUREMENTS
Temperature of the solution
Electrode surface area
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Electrode spacing
Concentration of the solution
THE DIFFERENCE BETWEEN STRENGTH AND CONCENTRATION OF AN
ACID OR A BASE
The strength of an acid or tells us how easily it ionizes to produce hydrogen ions while
concentration of an acid indicates the proportion of water water and acid present in aqueous
solution.
The strength of a base indicates how easily it ionizes to produce hydroxide ions while
concentration tells us the proportion of water and base present in aqueous solution.
OXIDES
Oxides are compounds containing oxygen and another element.
TYPES OF OXIDES
Oxides can are classified as basic, acidic or amphoteric.
1. BASIC OXIDES
These are formed when oxygen react with metals. Examples are magnesium oxide
(MgO), iron (II) oxide (Fe
2
O
3
) and copper oxide (CuO) etc.
These compounds are usually ionic in nature.
Most of them dissolve in water to form alkalis. Others are insoluble in water.
They can neutralize an acid.
2. ACIDIC OXIDES
These are formed when oxygen react with non-metals. Examples are carbon dioxide
(CO
2
), sulphur dioxide (SO
2
) and phosphorous pentaoxide (P
2
O
5
). They form carbonic
acid, sulphurous acid and phosphoric acid.
When dissolve in water, they form a solution that turns blue litmus paper red.
3. AMPHOTERIC OXIDES
These oxides have both acidic and basic properties.
They react with both bases and acids to form salt and water.
Examples include aluminium oxide (Al
2
O
3
), zinc oxide (ZnO) and lead (II) oxide (PbO)
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WAYS OF REGULATING pH IN DIFFERENT ENVIRONMENTS
The human digestive system, soils and water are some environments whose pH must be
maintained for them to function efficiently.
a. THE HUMAN DIGESTIVE SYSTEM
The hydrochloric acid present in the stomach helps in protein digestion. If excess
hydrochloric acid is produced, it causes the stomach to ache. To deal with this problem, an
antacid containing magnesium hydroxide solution (milk of magnesia) is taken in order to
neutralize the acid.
b. SOIL AND WATER ACIDITY
Acid levels in the soil increase when fertilizers are used for a long time. When acidic gases
dissolve in rain water, they make water more acidic. High acidity in water and soils can be
controlled by adding lime (calcium hydroxide).
WAYS OF PREPARING SALTS
A salt is product formed when the hydrogen ion of an acid is replaced by a metal or
ammonium radical. Salts that do not dissolve in water are said to be soluble salts while those
that do not dissolve in water are said to be insoluble salts. The method chosen for the
preparation of a particular salt depends on the solubility of the salt in water. The methods of
preparing salts include the following:
Reaction of an acid with soluble bases.
Reaction of acids with metals
Reaction of acids with insoluble bases
Reaction of acids with carbonates
Precipitations
a. Reaction of acid with a soluble base (alkalis).
This method is used to prepare the salts of very reactive metals such as potassium or sodium.
Titration is used in this reaction because both reactants are in solution. The acid is slowly and
carefully added to a measured volume of alkali using a burette until the indicator, usually
phenolphthalein, changes colour.
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Once the end point is reached, the same volume of acid is added to the measured volume of
alkali, but this time without the indicator. The solution which is produced can then be
evaporated slowly to obtain pure dry crystals of the salt.
For example
hydrochloric acid + sodium hydroxide sodium chloride + water
HCl (aq) + NaOH (aq) NaCl (aq) + H
2
O (l)
b. Reaction of an acid with a metal
This method can only be used with less reactive metals such as magnesium, aluminium, zinc,
iron and tin. Excess metal is added to dilute acid. The metal dissolves in then acid forming a
salt solution. The salt solution formed is evaporated slowly to form a saturated solution of the
salt. For example
magnesium + nitric acid magnesium nitrate + hydrogen gas
Mg (s) + 2HNO
3
(aq) Mg(NO
3
)
2
(aq) + H
2
(g)
c. Reaction of acid with carbonate
This method is used with any metal carbonate and any acid, providing the salt is soluble. The
procedure is similar to that of an acid and a metal. For example, copper (II) carbonate would
be added in excess to dilute nitric acid to produce copper (II) nitrate solution.
copper (II) + nitric acid copper (II) + carbon + water
nitrate nitrate dioxide
CuCO
3
(s) + 2HNO
3
(aq) Cu(NO
3
)
2
(aq) + CO
2
(g) + H
2
O
(l)
d. Reaction of an acid with an insoluble base
This method can be used to prepare a salt of an unreactive metal, such as lead or copper. In
this case, it is possible to use direct reaction of the metal with an acid so the acid is
neutralized using particular metal oxide. The reactants are warmed. A typical example is the
neutralisation of sulphuric acid by copper (II) oxide to produce copper (II) sulphate.
sulphuric acid + copper (II) copper (II) + water
oxide sulphate
H
2
SO
4
(aq) + CuO (s) CuSO
4
(aq) + H
2
O (l)
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e. Precipitation
This method is used to prepare an insoluble salt. Solutions of the two chosen soluble
solutions are mixed. The insoluble salt formed can be filtered off, washed with distilled water
and dried. For example, to produce barium sulphate, barium chloride and sodium sulphate
can be used.
barium + sodium barium + sodium
chloride sulphate sulphate chloride
BaCl
2
(aq) + Na
2
SO
4
(aq) BaSO
4
(aq) + 2NaCl (aq)
SOLUBILITY RULES
Solubility rules guide the choice of a method to be used when preparing salts.
All potassium (K), sodium (Na) and ammonium (NH
4
) are soluble.
All nitrates and hydrogen carbonates (HCO
3
) are soluble.
All sulphates are soluble except lead sulphate (PbSO
4
) and barium sulphate (BaSO
4
).
Calcium sulphate (CaSO
4
) is sparingly soluble.
All chlorides are soluble except silver chloride (AgCl). Lead chloride (PbCl
2
) is only
soluble in hot water.
All carbonates are insoluble except potassium, sodium and ammonium carbonates.
APPPLICATIONS OF PRECIPITATIONS REACTIONS
a. Treatment of water for domestic use
Sodium carbonate can be used to treat water for domestic use. The toxic cations are
precipitated as insoluble lead carbonate or mercury carbonates which are then filtered off.
Sodium carbonate is also used to make hard water soft by precipitating the ions that cause
water hardness.
b. Treatment of industrial effluents
Effluents refer to liquid wastes discharged from a sewage system, a factory, a nuclear powers
station or an industrial plant.
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When treating industrial effluents, precipitation is carried out to remove toxic ions in the
water solution. Ions to be removed are reacted with other ions to produce insoluble solids that
can be removed by sedimentation or filtration.
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TOPIC 3 : OXIDATION AND REDUCTION
1. OXIDATION
a. It is the addition of oxygen to a substance.
Consider the reaction between magnesium (Mg) metal and oxygen.
magnesium + oxygen magnesium oxide
2Mg (s) + O
2
(g) 2MgO (s)
The magnesium has gained oxygen. We say that it has been oxidized.
b. It is removal of hydrogen from a substance.
For example, ethanol can be changed to ethanal by an oxidising agent potassium dichromate
(VI), in the presence of an acid.
The ethanol molecule has lost two hydrogen atoms to become ethanal. It has been oxidized.
c. It is the loss of electrons by a substance.
Consider a reaction between magnesium and oxygen to form magnesium oxide.
2Mg (s) + O
2
(g) 2MgO (s)
In this reaction, each magnesium atom loses two electrons. Therefore magnesium has been
oxidized.
The oxidation reaction equation will be written as:
Mg Mg
2+
+ 2e
-
During oxidation, electrons are lost, so they are written on the side of products of the
equation.
d. It is the increase in charge or oxidation number.
Consider the reaction between sodium and chlorine to form sodium chloride.
2Na (s) + Cl
2
(g) 2NaCl (s)
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Each sodium atom gives one electron to the chlorine atom and increases its charge form (0) to
(+1) according to the equation below.
Na Na
+
+ e
-
It is also correct to write: Na
e
Na
+
Since sodium has increased its charge from (0) to
(+1)
, it has been oxidized.
REDUCTION
a. It is the removal of oxygen from a substance.
For example when copper oxide is heated with carbon, the following reaction takes place.
2CuO
(s)
+ C
(s)
CO
(s)
+ 2Cu
(s)
Copper oxide is losing oxygen. It is being reduced.
b. It is the addition of hydrogen to a substance.
For example during the chemical equation for the formation of ammonia gas is shown below.
N
2
(g) + H
2
(g) NH
3
Nitrogen is gaining hydrogen. It is being reduced.
c. It is the gaining of electrons by a substance.
Consider a reaction between magnesium and oxygen to form magnesium oxide.
2Mg (s) + O
2
(g) 2MgO (s)
In this reaction, each oxygen atom gains two electrons. Therefore, oxygen is being reduced.
The reduction reaction equation will be written as:
O
2
+ 4e
-
2O
2-
During reduction, electrons are gained, so they are written on the reactants part of the
equation.
d. It is the decrease in charge or oxidation number.
Consider the reaction between aluminium and chlorine to form aluminium chloride.
2Li (s) + Cl
2
(g) 2LiCl (s)
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During the reaction, each chlorine atom receives one electron from the lithium atom and
decreases its charge according to the equation below.
Cl
2
+ 2e
-
2Cl
-
The charge on chlorine has decreased from (0) to (-1). Therefore, chlorine has been reduced
OXIDATION NUMBERS
An oxidation number is a whole number with a positive or negative sign of the charge on an
atom, ion or a compound. It is also called the charge on an atom, ion or a compound.
It indicates the amount of electrons lost or gained by an element during a chemical reaction.
For example Al
3+
has an oxidation number of (+3) and O
2-
has an oxidation number of (-2).
ASSIGNING OXIDATION NUMBERS
While it is now clear that oxidation numbers can be positive or negative, it is important to
note that the calculation of such numbers follows a set of rules. The rules are:
1. The oxidation number of an uncombined element is equal to zero. For example the
oxidation number of carbon, C = 0, Cl
2
= 0, Fe = 0 and H
2
= 0, etc.
2. The oxidation number of each hydrogen atom, H is +1.
3. The oxidation number of each oxygen atom, O is -2.
4. The algebraic sum of oxidation numbers of the elements in a compound is equal to zero.
For example, H
2
O = 0 means that:
2(H) + 1(O) = 2(+1) + 1(-2) = 2 + (-2) = 0.
5. The oxidation number of monatomic ion is equal to the charge on the ion. For example,
the oxidation numbers of ions of group I elements is +1, group II is +2, group III is +3,
group VII is -1, etc. That is why the oxidation number of Na
+
= +1, Al
3+
= +3, Cl
-
= -1, P
3-
= -3, etc.
6. For a polyatomic ion, the sum of the separate charges inside the ion must be equal to the
final charge on the ion itself. For example, the oxidation number of
-
3
NO
is -1.
Example 1
Work out the oxidation number of sodium (Na) and chlorine (Cl) in sodium chloride (NaCl).
Solution
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It means that Na + Cl = 0. (Rule number 4)
But Na is in group I and Cl is in group VII of the periodic table.
Therefore the oxidation numbers of Na = +1 and Cl = -1
Example 2
Work out the oxidation number of S in
2-
4
SO
.
Solution
2-
4
SO
is a polyatomic ion with a charge of -2. The oxidation number of each oxygen atom, O
= -2. Let the oxidation number of sulphur be y.
So, y + 4(O) = -2 (Rule number 6)
y + 4(-2) = -2
y – 8 = -2
y = +6
Therefore the oxidation number of S in
2-
4
SO
is +6.
Example 3
In the equation below, work out the oxidation number of manganese (Mn) in
-
4
MnO
and
state if it is being reduced or oxidized.
-
4
MnO
+ 8H
+
Mn
2+
+ 4H
2
O
Solution
-
4
MnO
is a polyatomic ion with a charge of -1 and the oxidation number of each oxygen
atom is -2.
Let the oxidation number of manganese (Mn) = z
So, z + 4(O) = -1 (Rule number 6)
z + 4(-2) = -1
z - 8 = -1
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z = -1 + 8
= +7
Therefore the oxidation number of Mn is +7. The oxidation number of Mn decreases from
(+7) to (+2), hence manganese (Mn) is being reduced.
HALF EQUATIONS IN REDOX REACTIONS
Consider the equation for the reaction between sodium (Na) and chlorine (Cl)
2Na (s) + Cl
2
(g) NaCl (s)
In this reaction the following has happened:
Each sodium atom has donated one electron to chlorine atom. Therefore sodium has been
oxidized according to the equation below.
Na – e
-
Na
+
………………………..………….[oxidation reaction]
Each chlorine atom has received one electron from sodium atom. Therefore, chlorine has
been reduced according to the equation below.
Cl
2
+ 2e
-
-
2Cl
………..…...……………………….[reduction reaction]
Reduction of chlorine is taking place side by side with the oxidation of sodium. Since
reduction and oxidation occur simultaneously, this reaction is called reduction-oxidation
reaction or in short redox reaction. Each of the reactions is called a half-reaction because
the electrons released by one reaction are accepted by another in order to complete the redox
reaction.
SPECTATOR IONS AND NET IONIC EQUATIONS
Spectator ion: This is an ion which does not take part in a chemical reaction.
Ionic equation: This is an equation in which the ions which are involved in a reaction are
shown. The equation could be an overall reaction or half reaction. In a net ionic equation
spectator ions are excluded.
The equation for the reaction between iron (Fe) and copper sulphate (CuSO
4
) solution is
shown below.
Fe (s) + CuSO
4
(aq) Cu (s) + FeSO
4
(aq)
Writing the equation in terms of the ions present in solution we get:
Page 21
Fe (s) + Cu
2+
(aq) +
2-
4
SO
(aq) Fe
2+
(aq) +
2-
4
SO (aq)
+ Cu (s)
In this equation, the sulphate ion,
2-
4
SO
(aq) is found on both sides of the equation and
nothing has changed on it. This means the
2-
4
SO
ion did not take part in the reaction. It is the
spectator ion. The ionic equation for the reaction between iron (Fe) and copper sulphate
should therefore be written as follows:
Fe (s) + Cu
2+
(aq) Fe
2+
(aq) + Cu (s)
BALANCING REDOX EQUATIONS
There are two rules for writing redox equations
1. In any ionic equation, the charges must balance. This means the sum of charges on the
left-hand side (LHS), must be equal to the sum of charges on the right-hand side (RHS) of
the equation.
This rule applies to both overall (full) reaction and half-reaction equations. Examples are
Equation Charges on LHS Charges on RHS
Zn (s) + Pb
2+
(aq) Zn
2+
(aq) + Pb (s) +2 +2
Br
2
(aq) +
2e
2
-
Br
(aq) -2 -2
Na (s) Na
+
(aq) + e
-
0 0
2. The number of electrons lost in the oxidation half reaction must be equal to the number of
electrons gained in the reduction half reaction.
For example, in the reaction between iron (Fe) and copper sulphate (CuSO
4
) solution, the
number of electrons lost in the oxidation half reaction is two according to the equation below:
Fe(s) Fe
2+
(aq) + 2e
-
……………………………....[oxidation half equation]
Therefore, the number of electrons gained in the reduction half reaction must also be two.
Cu
2+
(aq) + 2e
-
Cu (s)………...……………................... [reduction half equation]
WRITING OVERALL EQUATIONS FROM TWO HALF REACTION EQUATIONS
To write an overall equation from two half reaction equations, the steps to be followed are:
Page 22
1. Balance the two half equations separately.
2. Combine the two half equations by adding using the rules of algebra. The electrons
should cancel each other out.
Example 1
The following are half equations for the reaction between magnesium (Mg) and copper
sulphate (CuSO
4
) solution.
o
Mg
(s) Mg
2+
(aq) + 2e
-
Cu
2+
(aq) + 2e
-
o
Cu
(s)
Write a full equation for the reaction.
Solution
The oxidation half reaction equation is:
o
Mg
(s) Mg
2+
+ 2e
-
Charges 0 +2 + -2
The reduction half equation is: Cu
2+
(aq) + 2e
-
o
Cu
(s)
Charges 2+ + 2- 0
Each of the half equations is balanced. Each side has zero charge (Rule number 1). In the
reaction, magnesium (Mg) loses two electrons in an oxidation reaction, which are gained by
the copper ions (Cu
2+
) in a reduction reaction (Rule number 2).
Adding the two half equations we get:
o
Mg
(s) + Cu
2+
(aq) + 2e
-
Mg
2+
(aq) +
o
Cu
(s) + 2e
-
Cancelling out the electrons, the full equation is
o
Mg
(s) + Cu
2+
(aq) Mg
2+
(aq) +
o
Cu
(s)
Example 2
The half equations for the reaction between zinc (Zn) and silver nitrate (AgNO
3
) solution are:
0
Zn
(s) - 2e
-
Zn
2+
(aq)
Page 23
Ag
+
(aq) + e
-
o
Ag
(s)
Write an overall equation for the reaction.
Solution
Each side of the half equations has zero charge. Therefore each half equation is balanced.
However, zinc (Zn) loses two electrons in an oxidation reaction but silver gains one electron
in a reduction reaction.
To equalize the number of electrons lost and gained, the reduction half equation must be
doubled. Thus
0
Zn
(s) - 2e
-
Zn
2+
(aq)……….………………..[oxidation half equation]
2Ag
+
(aq) + 2e
-
2
o
Ag
(s)………....……………[reduction half equation]
Adding the two half equations we get:
0
Zn
(s) - 2e
-
+ 2Ag
+
(aq) + 2e
-
Zn
2+
(aq) + 2
o
Ag
(s)
Cancelling out the electrons, the overall equation is:
0
Zn
(s) + 2Ag
+
(aq) Zn
2+
(aq) + 2
o
Ag
(s)
OXIDISING AGENTS AND REDUCING AGENTS
The terms oxidising agent and reducing agent are used to describe reactants in reduction–
oxidation (redox) reactions.
An oxidising agent is a substance that causes oxidation to take place by accepting electrons
from another substance. When accepting the electrons, the oxidising agent is itself reduced.
A reducing agent is a substance that causes reduction to take place by donating electrons to
another substance. When donating the electrons, the reducing agent is itself oxidized.
Examples
Write the oxidation and reduction half equations and identify the oxidising agent and the
reducing agent in the following equations.
1. Cu (s) + 2Ag
+
(aq) Cu
2+
(aq) + 2Ag (s)
Page 24
Solution
a. The half equations are:
Cu (s) Cu
2+
(aq) + 2e
-
……………....[oxidation half equation]
2Ag
+
(aq) + 2e
-
2Ag (s)……… ……[reduction half equation]
b. Reducing agent: Cu (s), it has increased its charge from 0 to +2
Oxidising agent: Ag
+
(aq), it has decreased its charge from +1 to 0.
2. Cl
2
(aq) + 2
-
Br
(aq) 2
-
C l
(aq) + Br
2
(aq)
Solution
a. The half equations are:
2
-
Br
(aq) Br
2
(aq) + 2e
-
………..[oxidation half equation]
Cl
2
(aq) + 2e
-
2
-
C l
(aq)………….[reduction half equation]
b. The reducing agent is
-
Br
(aq) while the oxidising agent is
2
Cl
(g)
3. Zn (s) + Pb(NO
3
)
2
Zn(NO
3
)
2
+ Pb (s)
Solution
a. The half equations are:
Zn (s) Zn
2+
+ 2e
-
……………[oxidation half equation]
Pb
2+
(aq) +2e
-
Pb (s)………..…..[reduction half equation]
b. The reducing agent is Zn (s) while the oxidising agent is Pb
2+
(aq)
ORDER OF REACTIVITY OF METALS
The order of reactivity of metals can be based on three criteria: reactions of metals with
water, reaction of metals with acid, and metals displacement reactions.
a. REACTIONS OF METALS WITH WATER
Some metals such as potassium, sodium, calcium, magnesium, zinc, copper, gold, silver and
iron react with water to produce a basic solutions or oxides and hydrogen gas. When their
reactions are compared, the order of reactivity of metals is generated as shown below:
Page 25
b. REACTION WITH HYDROCHLORIC ACID
Most metals react with dilute acids to form a salt solution and hydrogen gas. The order of
reactivity of metals with hydrochloric acid is shown below:
Page 26
c. DISPLACEMENT REACTIONS OF METALS
A metal displacement reaction is the reaction in which one metal substitutes another metal
from a solution. A metal displacing the other is in solid form while the metal being displaced
is in aqueous state in solution.
Example
Iron (Fe) metal is placed in an aqueous solution of copper (II) sulphate [CuSO
4
(aq)] and the
experiment is left for about 5 minutes. Observations are made and recorded.
Observations
After 5 minutes, the following should have been observed
Iron (nail) began to dissolve.
A black-brown coating of copper metal formed on the nail.
The colour of the solution began to turn from blue to pale green.
Explanation
When the iron (Fe) was placed in the solution, it started to dissolve and form iron ions (Fe
2+
).
This is oxidation reaction:
Fe (s) Fe
2+
(aq) + 2e
-
The copper ions (Cu
2+
) present in the solution captured the free electrons. This is reduction
reaction
Cu
2+
(aq) + 2e
-
Cu (s)
But since oxidation and reduction occur simultaneously, iron (Fe) goes into the solution as
Fe
2+
ions to take the place of copper ions [Cu
2+
(aq)]. The copper ions (Cu
2+
) come out of the
solution as copper metal
0
[Cu (s)]
and coat on the iron nail.
Page 27
Therefore, in this reaction iron is displaces copper. Since iron can displace copper, we say
that iron (Fe) is more reactive than copper (Cu).
The colour of copper (II) sulphate is blue because of the presence of the copper ions. In this
reaction, the copper ions are constantly being removed from solution to form copper metal.
As time goes by, the concentration of copper ions decreases causing the change in the colour
of solution from blue to pale green.
Displacement series
The displacement series is the arrangement of metals in order of their relative chemical
reactivity. The displacement series is also called activity series, reactivity series, or redox
series.
The activity series is as follows:
THINGS TO NOTE ABOUT THE ACTIVITY SERIES
1. The most reactive metal is at the top while the least reactive metal is at the bottom.
2. The metals at the top of the series can displace all other metals below them from solutions
of their compounds by forcing them to accept electrons. For example, Potassium (K) can
displace all the other metals below. Gold cannot displace any of the metals in the list.
Page 28
3. The metals high on the series can easily donate electrons to metals lower down the series.
For this reason, the metals at the top are said to be the most reducing or most
electropositive.
THE ELECTROCHEMICAL SERIES
This is a list of half equations that shows the ability of one metal to donate electrons to other
metals in solution. The series is as follows.
K (s) - e
-
K
+
(aq)
Na (s) - e
-
Na
+
(aq)
Li (s) - e
-
Li
+
(aq)
Ca (s) - 2e
-
Ca
2+
(aq)
Mg (s) - 2e- Mg
2+
(aq)
Al (s) - 3e
-
Al
3+
(aq)
Zn (s) - 2e
-
Zn
2+
(aq)
Fe (s) - 2e
-
Fe
2+
(aq)
Sn (s) - 2e
-
Sn
2+
(aq)
Pb (s) - 2e
-
Pb
2+
(aq)
H
2
(g) - 2e
-
2H
+
(aq)
Cu (s) - 2e
-
Cu
2+
(aq)
Ag (s) - e
-
Ag
+
(aq)
Au (s) - 3e
-
Au
3+
(aq)
The working principle of the electrochemical series is the same as that of the displacement
series. Atoms of an element above can displace ions of an element below. Atoms of an
element below will not displace ions of an element above.
POSITIONS OF THE METALS IN THE ELECTROCHEMICAL SERIES
The position of the metals in the series is important in predicting a displacement reaction as
well as the rate of reaction.
a. Whether a displacement reaction will take place or not
A reaction will take place if the atom is of an element which is high in the series while the ion
is of an element which is lower in the series.
For example, a reaction takes place when zinc (Zn) metal is mixed with a solution of copper
(II) sulphate.
Page 29
Zn (s) + CuSO
4
(aq) ZnSO
4
(aq) + Cu (s)
However, if the atom is of an element which is lower in the series while the ion is of an
element high in the series, a reaction will not take place.
For example, a reaction does not take place if copper (Cu) metal is mixed with a solution of
zinc sulphate (ZnSO
4
).
Cu (s) + ZnSO
4
(aq) No reaction
b. The rate of reaction
As the gaps between the elements become bigger, the reactions occur easily and fast. As the
gaps decrease, the reactions occur slowly. For example
Sn (s) + CuSO
4
(aq) Slow reaction
Mg (s) + CuSO
4
(aq) Fast reaction
Experiment
Aim: To investigate the displacement reactions of metals
You are provided with four beakers, distilled water, a measuring cylinder, sand paper, and
solutions of copper sulphate, zinc sulphate, iron sulphate and magnesium sulphate. You are
also provided with pieces of copper, zinc, iron and magnesium metals.
a. Pour about 2 cm
3
of copper sulphate solution into each of the four beakers.
b. Clean the copper, zinc, iron and magnesium metals using sand paper.
c. Put a piece of each metal into each of the four beakers containing copper sulphate
solution.
d. Record the in the table below by indicating “Reaction” or “No reaction”.
e. Rinse the beakers with distilled water.
f. Repeat steps (a) to (e) using solutions of zinc sulphate, iron sulphate and magnesium
sulphate respectively.
Metal salt solution Metal
Copper Zinc Iron Magnesium
Copper sulphate
Zinc sulphate
Iron sulphate
Magnesium sulphate
Page 30
g. Use the results to arrange the metals in order of increasing reactivity.
HINT:
The metal that reacts with all the given solutions is the most reactive.
The metal that reacts with none of the given solutions is the least reactive.
The metal that reacts with all the given solutions except the solution of the most reactive
metal is the second-most reactive.
The metal that reacts with all the solutions except the solutions of the first two most-
reactive is the third-most reactive metal.
THE ELECTROCHEMICAL CELL
The electrochemical cell is a system in which metallic electrodes are placed into an
electrolyte where a chemical reaction uses or generates electricity.
Electrochemical cells which generate electric current are called voltaic or galvanic cells.
THE VOLTAIC CELLS
The voltaic cells consist of two separate compartments called half-cells containing electrolyte
solutions and electrodes that can be connected in a circuit. Two dissimilar metals (e.g. copper
and zinc) are dipped in an electrolyte. If the metals are connected by an external circuit, one
metal is reduced while the other metal is oxidized.
HOW THE VOLTAIC CELLS WORK
Electrons flow from the more reactive metal (anode) to the less reactive metal (cathode).
For example, from the left to the right in the figure above.
Page 31
Oxidation occurs at the anode. The name refers to the flow of anions towards it. For
example, in the above voltaic cell zinc atoms give up electrons and go into the solution as
ions. This is oxidation.
Zn (s)
2e
Zn
2+
(aq)
Reduction occurs at the cathode. The name refers to the flow of cations towards it. For
example, in the voltaic cell above copper (II) ions, Cu
2+
(aq) accept electrons and copper
is deposited. This is reduction.
Cu
2+
+
2e
Cu (s)
The overall reaction is as follows: Zn (s) + Cu
2+
Zn
2+
(aq) + Cu (s)
The salt bridge acts as a transfer medium that allows ions to flow through out without
allowing the different solutions to mix and react.
Since electricity is defined as the flow of electrons, we can say that the redox reaction is
giving out electrical energy.
CELL NOTATION OF GALAVANIC CELLS
The standard notation used for all voltaic cells is:
anode|anode.electrolyte||cathode.electro
lyte|cathode
For example
2+ 2+
Zn(s)|Zn (aq)||Cu (aq)|Cu(s)
where | = phase boundary (solid/aqueous)
|| = salt bridge
STANDARD ELECTRODE POTENTIALS
The cell has an Electromotive Force (EMF) which is defined as maximum potential
difference between two electrodes of a voltaic cell. EMF is the driving force of the cell
reaction or any redox reaction i.e. the higher the EMF the stronger the reaction. The EMF is
measured by a voltimeter, in volts.
Experiments show that each half reaction has its own special voltage, which is called the
electrode potential; it is the potential difference between an electrode and the solution in
which it is immersed.
The electrode potentials of metals are different due to differences in reactivity. Metals high in
the electrochemical series tend to have high electrode potentials due to a greater ability to
donate electrons, which also means a greater voltage.
Metals lower in the series are made to accept electrons from those higher up, causing them to
have a lower voltage. This idea leads to the development of the voltage series.
Page 32
THE VOLTAGE SERIES
The voltage is a list of standard electrode potentials for metals. It is written as follows:
K (s) -
1e
K
+
(aq) -2.92 volts
Na (s) -
1e
Na
+
(aq) -2.71 volts
Mg (s) -
2e
Mg
2+
(aq) -2.31 volts
Al (s) -
3e
Al
3+
(aq) -1.67 volts
Zn (s) -
2e
Zn
2+
(aq) -0.76 volts
Fe (s) -
2e
Fe
2+
(aq) -0.44 volts
Sn (s) -
2e
Sn
2+
(aq) -0.14 volts
Pb (s) -
2e
Pb
2+
(aq) -0.12 volts
H
2
(g) -
2e
2H
+
(aq) 0.00 volts
Cu (s) -
2e
Cu
2+
(aq) +0.34 volts
Ag (s) -
1e
Ag
+
(aq) +0.80 volts
USING THE VOLTAGE SERIES
The electromotive force (EMF) of the voltaic cell depends on the difference in the half
cell voltages.
EMF = Oxidation electrode potential + Reduction electrode potential
=
redu ction ox idatio n
E E
Metals far apart in the voltage series would give a larger EMF, while metals close
together would give a small EMF.
CALCULATING ELECTROMOTIVE FORCE (EMF) OF A CELL
Write the half equation for each process.
Look up the standard electrode potentials for each reaction.
Reverse the sign on the oxidation electrode potential.
Add the electrode potentials to get the EMF.
A negative sign on the EMF indicates a spontaneous reaction will occur between the two
metals. A negative sign indicates no reaction.
Example
Calculate the electromotive force of the electrochemical cell,
2+ 2+
Zn(s)|Zn (aq)||Cu (aq)|Cu(s)
.
Refer to the voltage series.
Page 33
Solution
Reduction half reaction :
2+
Cu (aq)+2e Cu(s)
0
E 0.34V
Oxidation half reaction :
2+
Zn(s) Zn (aq)+2e
0
E 0.76V
EMF =
redu ction ox idatio n
E E
= +0.34V – (-0.76) V
= +1.10 V
CORROSION AND RUSTING
Corrosion is the reaction which occurs when some metals react with atmospheric oxygen
and water.
Corrosion is an example of a displacement reaction which results in degradation of metals by
dissolving and wearing away. During corrosion, the metals get oxidized to form oxides. For
example,
4Fe (s) + 3O
2
(g) 2Fe
2
O
3
(s)
Rusting means the oxidation of iron. Rust is the name of the iron oxide compound formed
when iron corrodes (i.e. reacts with oxygen and water).
HOW RUSTING OCCURS
Iron reacts oxygen to form (brown) iron (III) oxide.
4Fe (s) + 3O
2
(g) 2Fe
2
O
3
(s)
At the same time, the iron (III) oxide reacts with water to form brown hydrated iron (III)
oxide.
2Fe
2
O
3
(s) + H
2
O (g) 2Fe
2
O
3
.H
2
O
Iron (III) oxide water vapour hydrated iron (III) oxide
The hydrated iron (III) oxide is known as rust. Rust is a soft, crumbly brown solid.
RUSTING AS A REDOX REACTION
Firstly, iron dissolves in water and ionizes according to the equation:
Fe (s) – 2e
-
Fe
2+
(aq)
Water and air capture the electrons and are reduced according to the equation:
2H
2
O(l) + O
2
(g) + 4e
-
4OH
-
(aq)
Page 34
The hydroxyl ions react with the iron, Fe
2+
(aq) ions and more water to form rust. The net
ionic equation for the reaction is:
2Fe
2+
(aq) + O
2
(g) 4Fe
3+
(aq) + 2O
2-
(aq)
CONDITIONS FOR RUSTING
There are two conditions necessary for rusting to take place.
Presence of oxygen
Presence of water
If either oxygen or water is not present, then rusting will not take place.
INVESTIGATING CONDITIONS FOR RUSTING
Materials: test tubes, test tube rack, iron nails, cooking oil, anhydrous calcium chloride, tap
water
Procedure:
a. The apparatus is set up as shown below.
b. The test tubes are left undisturbed for one week.
Expected results
The nail in test tube A will rust while those in test tubes B and C will not rust.
Explanation
In test tube A, rusting occurs because of the presence of both oxygen and water. The tap
water contains air, hence oxygen. All the necessary conditions for rusting are present in test
tube A. In test tube B, rusting will not occur because the boiled water does not contain
dissolved oxygen. The oil prevents any atmospheric oxygen from entering. So, only water is
Page 35
present. In test tube C, the anhydrous calcium chloride is used to remove water. Hence only
air is present. As a result, rusting will not occur.
PREVENTION OF RUSTING
Rusting can be prevented using the following methods.
Painting. It prevents water and oxygen coming into contact with metal.
Greasing and oiling. It prevents water and oxygen from interacting with the metal.
Alloying. An alloy is a mixture of metals. For example, stainless steel is made by mixing
iron with chromium, nickel and manganese. The alloy is corrosion resistant.
Galvanizing. This involves covering or coating iron with a molten layer of zinc. The zinc
reacts with water and oxygen in preference in preference to iron.
Even if the zinc is scratched, the iron does not rust. This is because zinc is higher in the
activity series. Zinc coated iron is called galvanized iron. In galvanizing, zinc is being
sacrificed to protect iron. Hence this type of protection is called sacrificial protection.
Coating iron with plastic.
Electroplating. This is the process of coating a metal with a thin layer of another metal by
using electrolysis.
Page 36
TOPIC 4 : ELECTROLYSIS
Electrolysis is the decomposition of a compound which is in molten or aqueous state using
electrical energy. The process of electrolysis is carried out in an electrochemical cell.
Electrolysis works for
ionic compounds that are melted or dissolved in water, so that the ions are free to move.
covalent compounds that form ions when they are dissolved in water.
COMPONENTS OF AN ELECTROCHEMICAL CELL
The electrochemical cell consists of a dry cell, an ammeter, bulb, graphite rods and an
electrolyte.
An electrolyte is a compound which when melted or when dissolved in water can
conduct electricity and get decomposed by it.
The graphite rods are called electrodes. The electrode attached to the positive terminal of
the battery is called the anode.
The electrode connected to the negative terminal of the cell is called the cathode.
Dry cells are used to supply direct current.
The bulb or ammeter is used to indicate that current is flowing in the circuit.
ELECTROLYSIS OF MOLTEN IONIC COMPOUNDS
Consider the electrolysis of molten lead bromide in the figure below
Page 37
When the switch is closed this what happens:
Electrons move from the negative terminal of the battery to the cathode.
In the liquid, ions move to the electrode of opposite charge. Lead ions (Pb
2+
) are attracted
to the cathode while bromide ions (Br
-
) are attracted to the anode. The moving ions carry
electric current.
At the cathode, lead ions each receive two electrons and are reduced according to the
equation: Pb
2+
(l) + 2e
-
Pb (l)
Lead collects on the electrode and eventually drops off it.
At the anode, bromide ions each gives up an electron and are oxidised according to the
equation: 2Br
-
(l) - 2e
-
Br
2
(g)
The bromide gas bubbles off. Combining the anode and cathode reactions we get a redox
reaction.
2+
Pb
(l) +
2Br
(l) Pb (l) + Br
2
(g)
So it must be noted that electrolysis decomposes a molten ionic compound. A metal is
obtained at the cathode, and a non-metal at the anode.
ELECTROLYSIS OF AQUEOUS IONIC COMPOUNDS
When a salt such as sodium chloride (NaCl) dissolves in water its ions become free to move.
So it can be electrolysed. However, the products are usually different from the products of the
electrolysis of molten ionic compounds. This is so because the water itself also produces
ions; hydrogen ions (H
+
) and hydroxide ions (OH )
. A tiny proportion of its molecules split
into ions.
Page 38
So in the aqueous solution, the H
+
and
OH
ions are from water compete with other ions to
receive or give up electrons. To predict which ion will win the following rules apply during
the electrolysis of aqueous ionic compound.
1. At the cathode, either a metal or hydrogen is formed.
If a metal is more reactive than hydrogen, its ions stay in solution. The H
+
ions accept
electrons, and hydrogen bubbles off.
If the metal is less reactive than hydrogen, its ions will accept the electrons. The metal
forms, leaving the H
+
ions in solution.
2. At the anode, a non-metal other than hydrogen is formed.
If a halide ion
(Cl , Br , I )
is present in sufficient concentration, it will give up
electrons more readily than the
OH
ions do. Molecules of chlorine, bromine and iodine.
If no halide ion is present or if the halide solution is dilute, the
OH
ions will give up
electrons and oxygen will be formed.
In either case, the ions are discharged selectively since only at each electrode only one type
of ions gets discharged. This is called preferential discharge.
DEMOSTRATING PREFERENTIAL DISCHARGE
a. Electrolysis of brine (concentrated solution of sodium chloride, NaCl (aq)).
Page 39
The ions present in solution are Na
+
,
Cl
from the salt, and H
+
and
OH
from the water. The
figure below shows the reactions at each electrode as the electrolysis proceeds.
At the cathode, the H
+
accepts electrons since hydrogen is less reactive than sodium. The
equation for the reaction is:
2H
+
(aq) +
2e
H
2
(g). This is reduction. The hydrogen gas bubbles off.
At the anode, the
Cl
ions give up electrons more readily than the
OH
ions do. The
equation for the reaction is:
2Cl
(aq)
2e
Cl
2
(g). This is oxidation. The chlorine gas bubbles off.
When the hydrogen and chlorine gases bubble off, Na
+
ions and
OH
ions remain behind, so
a solution of sodium hydroxide is formed.
b. Electrolysis of dilute sodium chloride
The same ions are present as before. But now the proportion of Na
+
and
Cl
ions is lower
since the solution is dilute.
At the cathode, hydrogen ions gain electrons according to the equation:
4H
+
(aq) +
4e
2H
2
(aq)
This is reduction. (The 4 electrons are there to balance the half-equation at the anode)
At the anode, the
OH
ions give up electrons, since the
Cl
ions are in low concentration.
Oxygen gas bubbles off according to the equation:
4OH
(aq)
4e
O
2
(g) + 2H
2
O (l)
Page 40
Instead of the dilute sodium chloride being decomposed, it is the water that gets decomposed.
In a sense this is electrolysis of water.
EFFECTS OF ELECTRODES ON ELECTROLYSIS
Electrolysis of a solution will give different results depending on electrodes. A comparison is
shown in the following a and b.
a. Electrolysis of copper sulphate aqueous copper (II) sulphate using inert electrodes.
Inert electrodes are the electrodes which allow passage of electric current into and out of
electrolyte without themselves getting decomposed. Examples are carbon graphite and
platinum.
At the cathode, copper ions get discharged.
2Cu
2+
(aq) +
4e
2Cu (s)
The copper coats the electrode
At the anode, oxygen bubbles off.
4OH
(aq)
4e
O
2
(g) + H
2
(g)
The blue colour of copper sulphate solution fades as the copper ions are discharged.
b. Electrolysis of copper aqueous copper (II) sulphate using copper electrodes
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At the cathode, copper is formed and coats the electrode.
Cu
2+
(aq) +
2e
Cu (s)
The anode dissolves, giving copper ions.
Cu (s)
2e
Cu
2+
(aq)
The copper ions move to the cathode to form copper. The colour of the solution does not
fade.
WRITING HALF-EQUATIONS FOR ELECTRODE REACTIONS
To write the correct half-equations for electrode equations, follow the following steps:
a. Name the ions present and the products. Write each half-equation correctly.
b. Remember, positive ions go to the cathode and negative ions go to the anode.
c. Write the correct symbol of the element that forms. For example, Cl
2
(not Cl).
d. The number of electrons in the equation should be the same as the total charge on the
ions in it.
Example
Write the anode and cathode half-equations for the electrolysis of molten magnesium
chloride.
Solution
The ions present in solution are magnesium ions, Mg
2+
and chloride ions,
Cl
ions.
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At the cathode
2Cl ( )l
2e
Cl
2
(g)
At the anode
Mg
2+
+
2e
Mg (l)
USES OF ELECTROLYSIS
1. EXTRACTION OF METALS
Less reactive metals can be extracted from their ores by spontaneous redox reactions. The
metal is extracted by dipping the more reactive metal into an aqueous solution of the metal to
be extracted. For example, to extract copper (Cu) from its compound, an iron metal can be
dipped into the aqueous solution of a compound of copper such as copper (II) sulphate. The
reaction between iron and aqueous copper sulphate will be:
Fe (s) + CuSO
4
(aq) Cu (s) + FeSO
4
(aq)
2. PURIFICATION OF METALS
To purify a metal, the mineral ore is dissolved to form an electrolyte and current is passed
through it. The metal is extracted as a deposit on the cathode.
PURIFICATION OF COPPER
The apparatus is connected as shown below.
The anode is made of impure copper. The cathode is pure copper. The electrolyte is dilute
copper (II) sulphate solution.
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The copper in the anode dissolves into the solution. This is oxidation reaction. The half
reaction equation at the anode is: Cu (s) Cu
2+
(aq) + 2e
-
The copper ions are reduced and deposited on the pure copper cathode. The half reaction
equation at the cathode is: Cu
2+
+ 2e
-
Cu (s)
The impurities do not dissolve; they just drop to the bottom of the container.
3. ELECTROPLATING
Electroplating is the process of coating a metal with a thin layer of another metal by using
electrolysis.
To electroplate a metal, the procedure is as follows.
First, the metal to be electroplated is thoroughly cleaned.
An electrolyte whose ions are required to be deposited is selected.
Direct current (dc) is used instead of alternating current (ac).
The metal to be electroplated must be made the cathode. Metals for coating must be made
the anode so that when they dissolve into the electrolyte, they replace the ions which are
being removed from the container.
ELECTROPLATING OF IRON WITH COPPER
The apparatus is set up as shown below
The switch is closed and the experiment is observed for about 10 minutes.
Observations are made and recorded.
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Observations
The iron nail will be coated with copper metal.
The copper metal in the anode will be eaten up.
Explanation
When the switch is closed, the copper metal in the anode dissolves and is oxidised as follows:
Cu (s) Cu
2+
(aq) + 2e
-
The copper ions (Cu
2+
) are attracted to the cathode. At the cathode, the copper ions are gain
electrons and are reduced to copper atoms.
Cu
2+
(aq) + 2e
-
Cu (s)
The copper metal coats the iron metal. The iron metal is said to be electroplated.
APPLICATIONS OF ELECTROPLATING AND ITS BENEFITS TO SOCIETY
The process of electroplating is used:
To make objects look attractive. For example most necklaces and wristwatches are plated
with silver to make them look shiny and beautiful.
To protect metals from corrosion. For example, steel car bumpers are coated with
chromium to protect them from reacting with water.
To increase the strength or size of the metal. If a metal bar is small, it can be made bigger
by adding a layer on it through electroplating.
To extract and purify metals from ores e.g. copper.
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TOPIC 5 : ISOMERISM
Isomerism is the existence of compounds with the same molecular formula but different
structural formulae.
ISOMERS
Isomers are the compounds having the same molecular formula but different structural
formulae. Examples are
CONFORMERS
Conformations are compounds that differ from each other by the rotation of a single bond in
the molecule.
All the compounds above have the same molecular formulae and same structural formulae,
but they have different shapes due to bending, turning, or twisting of the carbon – carbon
bonds. Bending, turning or twisting of the C – C bonds in the molecule does not change the
structure of the molecule.
Page 46
IDENTIFYING ISOMERS OR CONFORMERS
In order to decide whether given structures are isomers or conformation of each other, it is
useful to check if there are any branches or junctions in the molecules the molecules have
simply been turned, twisted or bent.
To do this properly, the following steps may assist.
a. Working out the carbon skeleton first. The carbon skeleton enables us to see important
point about the structure of the molecule.
b. Choosing and numbering the longest chain of carbon atoms in the molecule. This chain
acts like a reference line inside the molecule. It is called the parent chain. The parent
chain allows us to locate accurately the number and position of the junctions in a
molecule.
c. Naming the compounds. Isomers have the same molecular formulae but different names
while conformations have the same molecular formulae and names.
For example, the following carbon structures represent conformations, because there are no
junctions in the molecules.
All of the conformations above have the same name: normal hexane (n – hexane).
The above compounds will have different names because they have different structural
formulae even though they have the same molecular.
TYPES OF STRUCTURAL ISOMERISM
There two types of structural isomerism; chain and position isomerism.
a. Chain isomerism
Page 47
This is the isomerism due to the differences in the way the carbon atoms are connected to
each other in the molecule. For example
b. Position isomerism
This is the isomerism due to the difference in the position of a functional group in the
molecule. For example
NAMING ISOMERS
Isomers are named systematically. The rules for naming the isomers of compounds vary from
one homologous series depending on whether the compounds in the series show chain
isomerism or position isomerism or both.
RULES FOR NAMING ISOMERS OF ALKANES
Alkanes do not have a functional group, so they only show chain isomerism. The rules for
naming the isomers of alkanes are as follows:
a. Write down the carbon skeleton for each molecule. Using
5 12
C H
, the carbon skeletons
are:
b. Number the parent chain in the carbon skeleton. Numbering starts the end of the parent
chain nearest the first branch.
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c. Count up all the carbon in the parent chain and use the number to give a basic name to the
compound.
d. Name the group that is attached to the parent chain. This group is called an alkyl group.
Alkyl group Name of the alkyl group
CH
3
- Methyl
CH
2
CH
3
Ethyl
CH
2
CH
2
CH
3
Propyl
CH
2
CH
2
CH
2
CH
3
Butyl
For straight chain single bonded carbon – carbon isomers, there is no branched group. Such
isomers start with the prefix ‘n-’ which stands for normal.
e. State the positions of the carbon atom on the parent chain where the branch (es) is/are
occurring and give a final name to the compound. The number is written before the alkyl
name. For example
If more than one alkyl group is attached to the parent chain, the numbers indicating the
positions of the branches are also written before the alkyl group. The numbers of the alkyl
Page 49
groups having the same number of carbon atoms are indicated by prefixes di-, tri-, or tetra- to
mean two, three or four respectively. For example
f. Name the compound fully.
RULES FOR NAMING ISOMERS OF ALKENES
Alkenes show both chain and position isomerism due to the presence of the functional group,
the carbon – carbon double bond in their structures. The procedure for naming isomers of
alkenes is as follows:
a. Write the carbon skeleton using the number of carbon atoms given in the molecular
formula. Place the functional group and branches in correct positions.
Let us consider the compound whose molecular formula is C
5
H
10.
b. Choose the parent chain and number it. Numbering starts at the end of the parent chain
nearest the functional group.
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c. Count up the number of carbon atoms in the parent chain and give a basic name to the
compound.
d. Show the position of the functional group by the lowest number of the carbon atom
holding the functional group. Put it before –ene. For example pent-1-ene or hex-2-ene.
e. State the position and type of the alkyl group attached. Name the compound fully.
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RULES FOR NAMING ISOMERS OF ALKANOLS
Alkanols contain the –OH functional group. For this reason, alkanols, just like the alkenes,
show both chain and position isomerism.
The rules for naming isomers of alkanols are as follows:
a. Draw a straight chain carbon skeleton using the number of carbon atoms and give its
name.
b. Place the functional group and branches in correct positions.
c. Number the parent chain. It is numbered in such a way that the carbon atom to which the
–OH group is attached has the lowest number.
d. Indicate the position of the functional group using the number of the carbon atom to
which the –OH group is attached. Put it before –ol. For example butan-2-ol.
Example
Write down and name all the isomers of the pentanol (C
5
H
11
OH).
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RULES FOR NAMING ISOMERS OF ALKANALS
Alkanals are compounds with a functional group called the carbonyl group, -CO. Alkanals
only exhibit chain isomerism. They do not show position isomerism because their functional
group is always placed at the end of the alkanal molecule.
Example 1:
Write down all the structural isomers of butanal
3 7
(C H CHO)
Solution
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Example 2:
Write down and name all the structural isomers of pentanal
4 9
(C H CHO)
.
Solution:
RULES FOR NAMING ISOMERS OF ALKANONES
Alkanones have a functional group similar to that of alkanals, the carboxyl group, -CO.
Alkanones exhibit both chain and position isomerism. This is because the carboxyl group in
alkanones can be shifted to different positions in the molecule.
The position of the functional group is indicated by the number of the carbon atom in the
carboxyl group. It is put before –one. For example, butan-2-one, pentan-3-one, etc.
Example
Write down and name all the structural isomers of pentanone
5 10
(C H O)
.
Solution:
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RULES FOR NAMING ISOMERS OF ALKANOIC ACIDS
Alkanoic acids are organic compounds whose functional group is the carboxyl group, –
COOH. Alkanoic acids only exhibit chain isomerism.
Example 1
Write down and name all the structural isomers of pentanoic acid (C
4
H
9
COOH).
Solution:
Page 55
TOPIC 6 : POLYMERISATION
Polymerisation is a chemical reaction in which small molecules join together to form long
chain molecules called polymers. Polymers are long – chain molecules made by joining
many small molecules called monomers. Monomers are the small molecules which join
together to form polymers.
TYPES OF POLYMERS
There are two types of polymers: natural polymers and synthetic polymers.
a. Natural polymers
These are polymers found in all living things. Examples of natural polymers are proteins,
carbohydrates, starch, wood, rubber and deoxyribonucleic acid (DNA).
b. Synthetic polymers
Synthetic polymers are manufactured in industries. They are normally called plastics.
Examples of synthetic polymers are: Polythene, Polyvinylchloride (PVC), Terylene, Nylon,
Perspex (artificial glass), Polytetraflouroethene (PTFE) or Teflon and Polyphenylethene
(Polystyrene).
TYPES OF POLYMERISATION
a. Addition polymerisation
This is a process which involves repeated addition reactions of monomers of the same type.
The monomers used in addition polymerisation are unsaturated compounds, for example
ethene or derivatives of alkenes e.g. chloroethene.
Examples of addition polymerisation are
The polymerisation of ethene
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If n molecules of ethene combine, the process may be written as follows:
HOW THE POLYMERISATION OF ETHENE OCCURS
During the polymerisation of ethene, ethene undergoes relatively high temperature and high
pressure in the presence of a catalyst. The double bonds between the carbon atoms are
broken, enabling many molecules of ethene to join to each other to form polyethene.
The polymerisation of chloroethene
b. Condensation polymerisation
This is a reaction whereby a small molecule is eliminated between two different monomers.
During condensation polymerisation:
two different monomers join.
the monomers join by releasing small molecules e.g. water or hydrogen chloride
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The best example of condensation polymerisation is the esterification reaction.
Carboxylic acid + Alkanol Ester + water
(Monomer 1) (Monomer 2) (Polymer)
Examples of condensation polymers are: Nylon, Terylene (polyester), etc.
PROPERTIES OF SYNTHETIC POLYMERS
Different polymers have unique characteristics. However, most of them will have the
following characteristics.
They do not corrode in air or water.
They are excellent heat insulators.
They are excellent electrical insulators.
They do not decompose easily.
They are relatively light.
Some are transparent
USES OF SYNTHETIC POLYMERS
Polymer Monomer Uses
Polyethene Ethene Used in making:
pipes and plastic bags
bowls and bottles
packaging material
insulation for electrical wiring
Polyvinylchloride (PVC) Chloroethene Used in making:
crates and bottles
plastic ropes
artificial leather
water pipes
insulation for electrical wiring
Polystyrene Styrene Used in making:
insulation material
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packaging (foam)
radio cabinets
pot handles
Polytetraflouroethene (PTFE)
or Teflon
Tetraflouroethene Used in making:
soles of iron
non-stick frying pans
Perspex
used as a glass substitute
used in lenses
ADVANTAGES OF SYNTHETIC POLYMERS
They are cheap to produce.
They are resistant to corrosion from acids, alkalis, water and air.
They are light and therefore portable.
They are soft and easy to manufacture.
They can be moulded into complicated shapes more easily.
They are good heat insulators.
They are good electrical insulators.
DISADVANTAGES OF SYNTHETIC POLYMERS
They are non-biodegradable hence difficult to dispose.
Some give off toxic gases such as carbon mono oxide when they burn.
They are expensive to recycle.
Some catch fire easily.
TYPES OF PLASTICS
There are two types of plastics: thermoplastics and thermosetting plastics.
a. Thermoplastics (thermosoftening plastics)
These are plastics that soften when heated and harden into a new shape when cooled. This is
because the intermolecular forces are weak and there are no cross-links or branches between
the polymer chains.
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Examples of thermoplastics are polythene, PVC, polystyrene, etc.
PROPERTIES OF THERMOPLASTICS
They are flexible. For this reason they do not break easily.
Their polymer chains are not cross-linked. They can be moulded into different shapes
several times.
They have low melting points. This is because they have weak intermolecular forces
between the polymer chains.
They can easily be recycled.
b. THERMOSETTING PLASTICS (THERMOSETS)
These are plastics which set hard and cannot be melted again once they are made.
This is because the polymers chains are held more strongly due to the presence of cross links
between the polymer chains.
Examples of thermosetting plastics are melamine, Formica, Bakelite, among others.
PROPERTIES OF THERMOSETTING PLASTICS
They are rigid. For this reason, they break rather than stretch under tension.
Their polymer chains are cross-linked. Therefore, they cannot be moulded into new
shapes after they are made.
They do not melt, but decompose on heating.
They cannot be recycled.
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DIFFERENCE BETWEEN THERMOPLASTICS AND THERMOSETTING
PLASTICS
Thermoplastics Thermosets
They soften and melt when heated They do not soften on heating
They can be re-moulded into new shape
after they are made
They cannot be re-moulded into new shape
after they are made.
They have no cross-links between polymer
chains
They have cross-links between polymer
chains
They can be recycled They cannot be recycled
Page 61
TOPIC 7 : WATER
NATURAL SOURCES OF WATER
The major natural sources of water can be classified into underground, surface and
atmospheric water sources.
1. Underground water sources
These are sources of water found in water tables. They include:
a. Wells. These are holes dug in the ground in order to obtain water.
b. Boreholes. These are deep holes drilled in the ground using machines. They lift water
using submersible pumps.
c. Springs of water. These are sources of water that flow out of the ground as small streams
or pools.
2. Surface water sources
Examples of surface water sources include dams, lakes, streams, rivers, seas and oceans.
3. Atmospheric water sources
These sources include rainwater collected from rooftops during rainy season.
THE WATER CYCLE
The water cycle is also called the hydrological cycle. It describes how water circulates
continuously between the earth’s surface and the atmosphere.
Heat from the sun causes evaporation from oceans, seas and lakes. It also causes
transpiration. Both evaporation and transpiration cause water vapour to form.
The water vapour rises, cools and condenses to form tiny droplets of water. These
droplets from clouds.
The clouds are moved by along by air currents. As they cool, the tiny droplets from larger
droplets which fall as rain when they reach a certain size.
The water that falls as rain runs into streams and rivers and then into lakes, seas and
oceans.
Page 62
PHYSICAL PROPERTIES OF WATER
Pure water is a colourless, tasteless and odourless liquid.
It boils at 100ºC and freezes and 0ºC.
It has a density of 1g/cm
3
.
IMPORTANCE OF WATER
Water is used:
a. as a coolant. In car engines it is used to maintain low temperature while in simple
distillation it is facilitates liquefaction process.
b. as a solvent. It is considered as universal solvent because it dissolves many things.
c. as a chemical reactant. It is use in many chemical reactions which occur in aqueous
media.
d. in hydroelectric power generation. Moving water is used to generate electricity.
e. for recreational purposes such as in swimming pools.
f. as a transport medium for ships, ferries, boats and canoes when transporting people and
goods.
HARD AND SOFT WATER
Page 63
Soft water is the water that contains low mineral content and produces soapy bubbles easily.
Examples of soft water are rainwater and distilled water. Hard water is the water that has a
high mineral content and does produce soapy bubbles easily. An example of hard water is tap
water.
TYPES OF WATER HARDNESS
a. Temporary hardness
Temporary hardness is caused by the presence of dissolved calcium or magnesium hydrogen
carbonates. Temporary hardness can easily be removed by boiling.
b. Permanent hardness
Temporary harness is caused by the presence of calcium or magnesium sulphates. Permanent
hardness is difficult to remove and cannot be removed by boiling. It can be removed by
distillation or chemical treatment.
ADVANTAGES OF HARD WATER
Calcium ions present in hard water are required by the body for bones and teeth.
Coates lead pipes with a thin layer of lead sulphate or lead carbonate and prevents the
lead from dissolving in water.
It is good for the formation of shells in animals.
Calcium and magnesium present in hard water are necessary for growth of both plants
and animals.
DISADVANTAGES OF HARD WATER
It wastes soap due to scum formation.
Causes electrical appliances such as kettles and boilers to fur hence making them
inefficient.
Can cause hot water pipes to block.
Spoils the finish of some fabrics.
METHODS OF REMOVING WATER HARDNESS
Removal of water hardness is based on whether the hardness is temporary or permanent. The
method and can be physical or chemical.
Page 64
1. REMOVING TEMPORARY HARDNESS
a. Boiling
When heated the hydrogen carbonates decompose to form insoluble carbonates. For example
b. Addition of calcium hydroxide or sodium hydroxide
Addition of calcium or sodium hydroxide to water causes precipitation of calcium and
magnesium ions in form of carbonates. For example
2. REMOVING PERMANENT HARDNESS
a. Distillation
The water is distilled away from the dissolved substances. This method is expensive to be
used on a large scale.
b. Addition of sodium carbonate (soda)
Calcium and magnesium ions are removed as precipitates. For example
c. Ion exchange
The hard water is passed through a container filled with suitable resin containing sodium
ions. The calcium or magnesium ions causing water hardness are exchanged for the sodium
ions in the resin. Sodium ions do not cause water harness.
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Sodium ions become part of the softened water and the calcium or magnesium ions become
part of the resin.
WATER POLLUTION
It is defined as the addition of harmful substances into water as result of human activities.
The harmful substances include toxic metals, plastics, pesticides, and fertilizers.
SOURCES OF WATER POLLUTION
a. Domestic sources.
This involves the release of human faeces and sewage into water bodies such as lakes and
rivers.
b. Industrial wastes
These include effluents (liquid wastes) originate from breweries, tanneries, textiles and paper
industries. The effluents contain toxic chemicals.
c. Agricultural wastes
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These include excessive pesticides, herbicides, weedicides, and fertilizers. These may enter
into rivers and lakes through run off after rains.
EFFECTS OF WATER POLLUTION
It can cause diseases such as cholera, typhoid and hepatitis.
It leads to destruction of ecosystems.
It can severely affect aquatic life by decreasing amount of oxygen in water bodies.
It can cause disruptions in food chains. This happens when toxins and pollutants in the
water are consumed by aquatic animals which are then consumed by humans.
WAYS OF CONTROLLING WATER POLLUTION
Avoiding excessive use of chemicals.
Treatment of sewage before releasing it into water bodies. It ensures that the pollutants
and disease-causing micro-organisms are destroyed.
Encouraging farmers to use farmyard manure instead of chemical fertilizers.
Constructing pit latrines away from water sources such as wells to avoid contamination of
the water source. The slope of the land should also be considered in putting up a latrine.
Proper legislation should be put by government to reduce improper damping of industrial
and domestic wastes.
Petroleum companies should put measures in place to avoid unnecessary accidents
resulting from spillage of oils into water bodies.
Page 67
TOPIC 8: GREENHOUSE GASES AND THE OZONE LAYER
NATURAL DISASTERS
Natural disasters are extreme, sudden events caused by environmental factors that injure
people and damage property. Examples of natural disasters are drought, windstorm,
earthquake, wild fires, floods, hurricanes, thunderstorms and volcanic eruptions.
IMPACT OF HUMAN ACTIVITIES ON NATURAL DISASTERS
Human activities can have an impact on natural disasters through floods following these
activities;
Deforestation. Hard surfaces such as sewers, parking lots, and concrete surfaces
causes water to move faster as compared to natural channels. Cutting down natural
vegetation and trees leaves land bare resulting in more erosion when water flows.
Urban development. Urban development has an impact on natural disasters. It
destructs the natural system of waterways.
Natural wetlands destruction. The destruction of natural wetlands causes devastating
floods during peak floods along river basins. More natural wetlands lower water level
during heavy rains.
Agriculture. Large scale practice of agriculture can result in the depletion of top soils
causing weakening of vegetation and inability to protect against flooding.
EFFECTS OF NATURAL AND HUMAN ACTIVITIES ON THE ATMOSPHERE
Destruction of buildings
Death of animals and human beings
Change in landscape
Damage of vegetation
Global warming due to production of harmful gases
GREENHOUSE GASES
Greenhouse gases are components of air whose accumulation in the atmosphere cause rise in
global temperatures. Examples of greenhouse gases are carbon dioxide, sulphur dioxide,
nitrogen oxides, and methane.
Page 68
SOURCES OF GREENHOUSE GASES
Greenhouse gas Sources
Carbon dioxide (CO
2
)
Combustion
Respiration
Fermentation
Decomposition
Methane (CH
4
)
It is the primary component of natural gas.
Wetlands
Oceans
Activities of termites
Nitrogen oxides (NO, NO
2
)
Natural processes e.g. lightning and biological
processes that occur in plants and animals.
Combustion of fossil fuels in car engines and in
industries.
Sulphur dioxide (SO
2
) Sources include fossil fuels combustion at power
plants and extraction of metal ores
GLOBAL WARMING
Global warming is the continued rise in the average temperature of the earth due to the
accumulative effect of the greenhouse gases in the atmosphere.
HOW DO GREENHOUSES CAUSE GLOBAL WARMING?
The sun sends out energy as light and ultraviolet (UV) rays.
These warm the earth, which reflect some of the energy away again, as heat.
Some of this heat escapes from the atmosphere. However, some of reflected heat energy
is absorbed by greenhouse gases in the atmosphere. So, the air and the earth are warmed.
CONTRIBUTION OF CARBON DIOXIDE AND METHANE TO GLOBAL
WARMING
Two main greenhouse gases believed to be responsible for global warming are carbon
dioxide and methane. The levels of both gases are rising.
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The level of carbon dioxide is rising because we burn more fossil fuels each year. The
carbon dioxide from this goes into the atmosphere.
The level of methane is increasing because there is an increase in animal farming and rice
farming around the world, and more landfill sites.
WAYS OF MITIGATING EFFECTS OF GLOBAL WARMING
Use of renewable sources of energy such as solar energy, wind power, and hydroelectric
power.
Automobiles can be fitted with catalytic converters to clean up exhaust gases before they
are released into the atmosphere.
Afforestation and deforestation in order to reduce the carbon dioxide levels in the
atmosphere.
AIR QUALITY STANDARDS
Internationally there are standards of air quality which limit emissions of gases. The air
quality standards being used currently are based on the World Health Organisation (WHO.)
In line with the WHO guidelines, the Malawi government through the Malawi Bureau of
standards (MBS) published guidelines on maximum allowable emissions of both gases and
particulate pollutant.
The table below shows ambient air quality standard limits in Malawi.
Pollutant Maximum
concentration in
ambient air
Averaging period
Suspended particulate matter 25µg/m
3
1 day
Carbon monoxide
9 ppm 8 hours
35 ppm 1 hour
Sulphur dioxide
0.20 ppm 1 hour
0.08 ppm 1 day
0.02 ppm 1 year
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Nitrogen dioxide 0.3 ppm 1 year
Ozone 0.12 ppm 1 hour
Lead 0.50µg/m
3
1 year
Photochemical oxidants 0.10 ppm 1 hour
0.08 4 hours
The table below shows air quality standard for motor vehicles in Malawi.
Pollutant Maximum allowable
concentration
Duration Typical high concentration
for specific pollutant, ppm
Hydrocarbons 160µg/m
3
3 hours 6
Carbon monoxide 10µg/m
3
8 hours 40
Nitrogen dioxide 100µg/m
3
1 year 0.1
Oxidants 160µg/m
3
1 hour 0.6
ppm means parts per million
µg/m
3
means concentration in micrograms per unit volume
THE OZONE LAYER
The ozone refers to a region of the earth’s stratosphere that absorbs most the sun’s ultraviolet
radiation. It is mainly found in the lower parts of the stratosphere from approximately 20km
to 30km above the earth.
IMPORTANCE OF THE OZONE LAYER
The ozone layer acts a s natural ultraviolet (UV) rays shield for the earth. It absorbs 93% to
99% of the sun’s ultraviolet light. The UV light is very dangerous to life on earth.
Exposure to UV rays may cause:
Skin cancer
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Damage to crops
Harm to aquatic life
DEPLETION OF THE OZONE LAYER
The ozone layer is destroyed by chlorofluorocarbons (CFCs). Things that release CFCs
include coolants, foams and aerosols. When CFCs get to atmosphere, the chlorine atom is
released from the CFC molecule. The released chlorine atom split the ozone to oxygen atoms
and oxygen molecules. The free oxygen atom then combines with ozone molecule to form
another oxygen molecule.
For example
O
3
(g) O
2
(g) + [O]
O
3
(g) + [O] 2O
2
(g)
The action of the CFCs results in the depletion of the ozone layer and reduces its UV natural
shielding capacity.
PROBLEMS ASSOCIATED WITH THE DEPLETION OF THE OZONE LAYER
Exposure to UV rays has several health related and environmental effects. Some of them are:
Increased risk for developing several types of skin cancer.
Cataract problems
Damage to immune system.
Aging of the skin. This results in a person looking older than what they actually are.
Respiratory problems such as difficulty in breathing, chest pain, and throat irritation.
Leads to the decline of some amphibians by affecting every stage of their life cycle.
Causes disruptions in the food chain and the oceanic and carbon cycle.
It alters physiological and developmental processes of the plants.
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TOPIC 9 : WASTE MANAGEMENT
Waste refers to unwanted materials produced by human activities.
CLASSIFICATION OF WASTES
Wastes can be classified based on different criteria; physical state and degradability.
CLASSIFICATION BASED ON PHYSICAL STATE
Based on physical state, wastes can be classified as solid wastes, liquids wastes, gaseous
wastes and sludge wastes
CLASSIFICATION BASED ON DEGRADABILITY
Based on degradability, wastes can be classified as:
Biodegradable wastes. These are wastes that can decay relatively quickly as a result of
the action of bacteria. They include animal and plant waste.
Non–biodegradable wastes. These are wastes that do not decompose naturally.
Examples include plastics and glass.
SOURCES OF WASTES
a. Domestic sources: These wastes include food leftovers, plastics, metals, untreated
sewage, expired or unused medicines and their containers.
b. Industrial sources: These include used oil, toxic gases, metals and plastics etc.
c. Agricultural sources: These include plant remains excess agricultural chemicals such as
pesticides, herbicides, and weedicides.
d. Medical sources: These are wastes from medical centres and hospitals. Examples are
plastics, glassware, expired drugs, used syringes, needles, cotton and detergents.
e. Laboratory sources: These include expired chemicals, contaminated chemicals, broken
glassware and detergents.
WAYS OF TREATING AND DISPOSING WASTE
a. Open burning
Wastes such as papers, plastics are usually burnt to ashes in open land. This method is
discouraged because it causes air pollution due to emission of harmful greenhouse gases.
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b. Incineration
This approach involves combustion of waste material. The refuse is burn to ashes in
controlled chambers called incinerators. Incineration quickly reduces waste volume, reduces
transportation costs and decreases emissions of harmful green house gases
c. Dump site
This is an open disposal site where wastes are deposited and left uncovered and uncontrolled.
It is not good for the environment as it is likely to cause air and water pollution.
d. Sanitary landfills
These are land disposal sites used to bury solid waste. The solid waste is spread in thin layers
compacted and covered with a fresh layer of soil such as clay. It is also characterized by the
absence of surface water bodies or low water table, preventing the risk of water pollution.
e. Compositing
This is the controlled aerobic decomposition of organic waste materials by the action of small
bacteria. The waste is dumped into a dug pit so that it decomposes. After it has decomposed,
it can be used by farmers as manure.
f. Recycling or re-use
Recycling involves reprocessing of materials that would be considered waste. For, most
plastics and paper can be recycled instead of burning them.
Advantages of recycling materials
It reduces pollution.
It reduces the cost of producing new products.
Re-using on the other hand involves using a material considered waste for another purpose
that was not initially intended for.
For example, instead of dumping used bottles and cans, we can use them for storing water in
our homes.
WAYS OF MINIMIIZNG WASTES
Waste can be minimized by three main ways. These are:
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Recycling
Reusing
Reduction at source
RECYCLING OF METALS AND PLASTICS
Recycling of most materials including metals and plastics proceed as follows:
a. Collection – getting the recyclable plastics from different waste disposal sites.
b. Sorting – the plastics are sorted out in terms of type, colour, even how it was made.
c. Cleaning – the goal is to remove impurities and everything that is not recyclable.
d. Resizing – it consists of shredding the material into small particles, making it easier to
process, reshape.
e. Identification – the small particles are tested to determine their quality and class.
f. Compounding – it involves smashing and melting the material pellets. The pellets can
then be used in the production of other products.
IMPORTANCE OF RECYCLING METALS AND PLASTICS
It preserves natural resources. It prevents depletion of the metals and materials used to
manufacture plastics.
It reduces emissions of harmful greenhouse gases.
It promotes economic development by saving money that could be spent on mining and
processing of new metals and plastics. The money can be channeled into other
development programs that can help raise the living standards of the people and the
economy as a whole.
It saves money since recycled materials are generally cheap. This means the actual cost of
buying the material is reduced hence saving money.
